\newproblem{lay:1_5_25}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 1.5.25}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	\begin{enumerate}[a]
		\item Suppose $\mathbf{p}$ is a solution of $A\mathbf{x}=\mathbf{b}$, so that $A\mathbf{p}=\mathbf{b}$. Let $\mathbf{v}_h$ be any
		      solution of the homogeneous equation $A\mathbf{v}_h=\mathbf{0}$ and let $\mathbf{w}=\mathbf{p}+\mathbf{v}_h$. Show that
					$\mathbf{w}$ is a solution of $A\mathbf{x}=\mathbf{b}$.
		\item Let $\mathbf{w}$ be a any solution of $A\mathbf{x}=\mathbf{b}$, and define $\mathbf{v}_h=\mathbf{w}-\mathbf{p}$. Show that
		      $\mathbf{v}_h$ is a solution of $A\mathbf{x}=\mathbf{0}$. This shows that every solution of $A\mathbf{x}=\mathbf{b}$ has the form
					$\mathbf{w}=\mathbf{p}+\mathbf{v}_h$ with $\mathbf{p}$ a particular solution of $A\mathbf{x}=\mathbf{b}$ and $\mathbf{v}_h$ a solution
					of $A\mathbf{x}=\mathbf{0}$.
	\end{enumerate}
}
{
  % Solution
	\begin{enumerate}[a]
		\item Let us check whether $A\mathbf{w}=\mathbf{b}$\\
		      $\begin{array}{rcll}
		        A\mathbf{w}&=&A(\mathbf{p}+\mathbf{v}_h) & \text{By definition of }\mathbf{w} \\
						           &=&A\mathbf{p}+A\mathbf{v}_h  & \text{By distributivy of matrix product} \\
											 &=&\mathbf{b}+\mathbf{0}     & \text{By definition of }\mathbf{p}\text{ and }\mathbf{v}_h\\
											 &=&\mathbf{b}                 & \text{Because }\mathbf{0}\text{ is the neutral of vector addition}
		      \end{array}$
		\item By definition of $\mathbf{w}$ and $\mathbf{p}$ we have
		      \begin{center}
						$A\mathbf{w}=\mathbf{b}$ \\
						$A\mathbf{p}=\mathbf{b}$ \\
		      \end{center}
					If we subtract both equations, we obtain
		      \begin{center}
						$A\mathbf{w}-A\mathbf{p}=\mathbf{b}-\mathbf{b}$ \\
						$A(\mathbf{w}-\mathbf{p})=\mathbf{0}$
		      \end{center}
					Taking into account that $\mathbf{v}_h=\mathbf{w}-\mathbf{p}$, this means
		      \begin{center}
						$A\mathbf{v}_h=\mathbf{0}$
		      \end{center}
					As required by the exercise, we have proven that $\mathbf{v}_h$ is a solution of the equation $A\mathbf{x}=\mathbf{0}$.
	\end{enumerate}
}
\useproblem{lay:1_5_25}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
